# Is function from topological group to metric space Borel?

Let $G$ be a pseudometrizable compact abelian topological group, $X$ a compact metric space and $f:X\rightarrow G$ a continuous bijective function.

Suppose there exists $g\in G$ such that if $d_{G}(g_{1},g_{2})\leq\epsilon$ then there exists $n$ such that $d_{X}(f^{-1}g^{n}g_{1},f^{-1}g^{n}g_{2})\leq\epsilon.$

If $G$ is not metrizable then in general $f^{-1}$ is not continuous but can we conclude $f^{-1}$ is Borel?

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Could you please add a few words of background and information? Otherwise it looks "too localized". –  Goldstern Apr 21 '13 at 17:36
Sure. The multiplication of g is a dynamical system, a rotation in a compact abelian group. I want to construct a dynamical isomorphism from a dynamical system in X to G. –  FelipeG Apr 22 '13 at 18:13
Suppose $G$ is pseudometrizable but not metrizable. The the closure of $\{e\}$, (the identity), is a closed subgroup $N$ of $G$. And every open set in $G$ either contains $N$ or is disjoint from $N$. Then this same thing is true for every Borel set. On the other hand, for any two points of $X$, there is an open set that contains one but not the other. So, whatever bijection $f$ we choose, it is not Borel.